WUFI

Hi -

I am analyzing the bottom section of a wall where it terminates at a roof. I'm including the roof in the model. All components have been modeled, but when I go to the "Grid" screen, I have an error message indicating "Layer #4 has thickness zero." I have reviewed all the geometry and everything is properly located and sized. The computations do not run when I get this error message. Does anyone know how to fix this problem?

Thanks.

I sent you a pm.
Christian

Hi

I am experiencing the same issue, have reviewed the geometry and cannot find an error. Any eplanation?

Hi,

the grid generator divides your geometry into horizontal and vertical lines. Dividing locations are the boundaries of your previously defined rectangles. Now it can happen that two rectangles are "almost" at the same position (x or y) and splitting at that position results in a thickness of zero.

This sometimes also happens - if you use values smaller than one mm - by rounding errors. We recommend not to use half millimeters (the influence in a building assembly is rather small).

You can find the "missing" layer by going to the grid section, then click in the table through the grid. The grid is marked with another color in your construction above. You will come to a line in the table with thickness is zero and the grid is not marked. So remember the layer before and after and between that two is your "missing" element.

Christian

Allright... I have previously made models with 0.1 mm as the smallest division without any problems. So this time I must have been unlucky. I will try and simplify the construction.

I am also getting this error. I understand that it is a rounding issue, but I was wondering if there is a way to make the program still run. Is there some way to either make the program have a different rounding tolerance, or a way to manipulate the surrounding geometry to accommodate.

We are working with a system that has a film that is .005" (0.127 mm). The funny thing is after messing with the geometry I found that the geometry to the left of the film and and the film itself would run just fine, but error after I placed geometry to the right of it.

This is kind of confusing, but I can provide the geometry to help show what I am talking about.

Hello Rasava,
we advice to implement films with a minimum thickness of 1mm. Change the diffusion vapor resistance according to the thickness of the layer.

For building issues it is usually not necessary/ reasonable to use layer thicknesses smaller than one mm.

If you are talking about a film on the outside/surface of your construction, you can consider it by setting a surface sd-value or vapor diffusion coefficient in the climate boundary condition.
Christian

HI,

When you say "Change the diffusion vapor resistance according to the thickness of the layer", do you mean that we have to extrapolate the diffusion vapor ressistance from the value of my film (0.127mm) to the diffusion vapor resistance of the layer 1mm (input for my model in WUFI2D)? is the extrapolation lineal?

thanks

Hi Alberto,

the vapor diffusion resistance of a layer is proportional to the resistivity of the layer material and proportional to the thickness of the layer. So you can use in WUFI a layer with, say, ten times the original thickness if you reduce the resistivity to one tenth. This modified layer will have the same vapor resistance as the original layer.

In WUFI (using the SI system), the resistivity of a material is specified by its µ-value. The µ-value states by how much the diffusion resistivity of the material in question is higher than that of stagnant air. The vapor diffusion resistance of a layer is given by its sd-value. The sd-value is the thickness of a layer of stagnant air which has the same vapor diffusion resistance as the layer in question. The sd-value of a layer is the µ-value of the layer material, multiplied by the layer thickness.

Example:

Suppose you have a vapor retarder membrane with thickness 0.1 mm. The membrane material has the µ-value 100 000. The sd-value of this membrane is therefore 0.0001 m * 100 000 = 10 m, it has the same vapor resistance as a 10 m thick layer of stagnant air.

In WUFI's assembly, it is not very practical to have a layer with thickness 0.1 mm, we recommend to use at least 1 mm. This modified layer is ten times thicker than the original layer. To have the same vapor resistance as originally, you have to reduce the resistivity of the material to one tenth of the original value, in this case the new µ-value is 10 000. The sd-value of the modified material is the same as before: 0.001 m * 10 000 = 10 m.

Kind regards,
Thomas

thanks